# Thread: Find function base on first order derivatives

1. ## Find function base on first order derivatives

Suppose that the function $\displaystyle f: \mathbb {R}^2 \rightarrow \mathbb {R}$ has the properties of:

a) $\displaystyle f(0,0)=1$
b) $\displaystyle \frac { \partial f }{ \partial x } (x,y) = 2$ and $\displaystyle \frac { \partial f }{ \partial y } (x,y) = 3 \ \ \ \ \ \forall (x,y) \in \mathbb {R}^2$

Prove that $\displaystyle f(x,y)=1+2x+3y$

Well, I guess I can integrate the derivatives and get this answer, but I have to solve it with methods of mean value theorem and gradients... How should I proceed? Thank you.

Suppose that the function $\displaystyle f: \mathbb {R}^2 \rightarrow \mathbb {R}$ has the properties of:
a) $\displaystyle f(0,0)=1$
b) $\displaystyle \frac { \partial f }{ \partial x } (x,y) = 2$ and $\displaystyle \frac { \partial f }{ \partial y } (x,y) = 3 \ \ \ \ \ \forall (x,y) \in \mathbb {R}^2$
Prove that $\displaystyle f(x,y)=1+2x+3y$
Start by looking at the function $\displaystyle g(x) = f(x,0)$. Apply the mean value theorem to show that $\displaystyle g(x) = 1 + 2x$. Then, knowing that $\displaystyle f(x,0) = 1+2x$, fix x and apply the mean value theorem to the function $\displaystyle h(y) = f(x,y)$ to show that $\displaystyle h(y) = h(0) + 3y = 1+2x+3y$.